Problem 17 - Calculus for Biology and Medicine

Question

Use Leibniz's rule to find \(\frac{d y}{d x}\). $$ y=\int_{0}^{1-4 x}\left(2 t^{2}+1\right) d t $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -12 + 64x - 128x^2 \).

Step 1: Identify Leibniz's Rule

Leibniz's rule for differentiation under the integral sign is applicable here, which states: if \( F(x) = \int_{a(x)}^{b(x)} f(t, x) \, dt \), then \( \frac{dF}{dx} = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(t, x) \, dt \). In this problem, \( a(x) = 0 \) and \( b(x) = 1 - 4x \). We will apply this rule to find \( \frac{dy}{dx} \).

Step 2: Compute partial derivatives and evaluate functions

First, evaluate the function at the upper limit: \( f(t) = 2t^2 + 1 \). So, \( f(b(x)) = f(1-4x) = 2(1-4x)^2 + 1 \). Compute the derivative of the upper limit \( b(x) \): \( b'(x) = \frac{d}{dx}(1-4x) = -4 \). Since \( f \) is independent of \( x \), \( \frac{\partial}{\partial x}f(t, x) = 0 \).

Step 3: Apply Leibniz's Rule

According to Leibniz's rule: \( \frac{dF}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) + 0 \). Implementations for the limits: \( b(x) = 1 - 4x \) and \( a(x) = 0 \) results in \( a'(x) = 0 \). The derivative then simplifies to: \[ \frac{dF}{dx} = [2(1-4x)^2 + 1](-4) \].

Step 4: Simplify the expression

Expand and simplify: \( 2(1-4x)^2 = 2[(1 - 8x + 16x^2)] = 2 - 16x + 32x^2 \). Add \( +1 \) to get \( 3 - 16x + 32x^2 \). Thus, the derivative becomes \(-4(3 - 16x + 32x^2) = -12 + 64x - 128x^2\).

Step 5: Final Step: Conclusion

The derivative of the original integral with respect to \( x \) is \( \frac{dy}{dx} = -12 + 64x - 128x^2 \).

Key Concepts

Differentiation Under the Integral Sign

In calculus, differentiation under the integral sign is a powerful technique that allows us to differentiate an integral whose limits or integrand depends on a variable. This method is particularly useful in solving complex integrals where the bounds change with a parameter. The classical tool for this task is Leibniz's rule, a formula that relates the derivative of an integral to the derivatives of its bounds and its integrand.

Leibniz's rule shows that when you have an integral of the form \( F(x) = \int_{a(x)}^{b(x)} f(t, x) \, dt \), differentiating it involves more than just taking the derivative of the integrand. The rule states:

• Calculate the effect of changing the upper and lower limits of integration.
• Consider the partial derivative of the integrand with respect to the variable.

Thus, the derivative \( \frac{dF}{dx} \) is given by this generalized formula:\[\frac{dF}{dx} = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(t, x) \, dt\]This method can simplify otherwise daunting calculus problems, turning them into more manageable expressions. Understanding each term’s interpretation helps in correctly applying Leibniz’s rule.

Calculus Problem Solving

When faced with a calculus problem involving integration, it's pivotal to break down the task into manageable steps. Calculus problem-solving requires understanding the problem, identifying applicable rules, and then proceeding with careful calculations.

For instance, in our exercise, you start with identifying that the integral involves variable limits, prompting the use of Leibniz’s rule. Once the rule’s applicable, you need to:

• Compute any required derivatives of the integrand and the limits of integration.
• Evaluate the function at those limits.
• Apply the rule and simplify the expressions obtained.

Remember, practice is key. Getting comfortable with these steps takes time, but with experience, problem-solving in calculus can become intuitive. Moreover, understanding why each step is necessary will help you learn not just how, but why, certain strategies are used. It represents the core of overcoming seemingly complex calculus problems.

Integrals with Variable Limits

Integrals with variable limits add an interesting layer of complexity to calculus problems. Unlike ordinary integrals, where limits are constants, variable limits depend on an external parameter, such as \( x \). This dependency introduces additional terms when differentiating the integral.

For example, in our problem, the upper limit is \( 1 - 4x \). This indicates that as \( x \) changes, the upper boundary of the region being integrated also shifts. Recognizing how these boundaries influence the result is critical in such problems. The differentiation process, governed by Leibniz's rule, includes terms that account for this movement:
The role of the derivative of the limits, like \( b'(x) \) in our case, must be understood as it alters the integral's output as \( x \) changes. Overall, acknowledging and analyzing how variable limits affect your calculus problems is essential for accurate solutions.